Let $X$ be simply connected and let $ \tau $ be a cycle in $X$. If $ \tau(0)=\tau(1)=x_0$ and $\gamma $ is a path between $x\in X$ and $x_0$, is $\tau$ homologous to $ \gamma\tau\gamma^{-1} $?

I think it is, but I'm not sure about the proof. Can I use the fact that $ \gamma\tau\gamma^{-1} $ is homologous to $ \gamma+\tau+\gamma^{-1} $, where $+$ is the operation between 1-chains (that is abelian)?


You can use that fact, and in fact you can even prove it !

Lemma: if $\alpha, \beta$ are two paths, $\alpha(1) = \beta(0)$ then $\alpha\beta$ is homologous to $\alpha + \beta$

To prove this, we need to find a $2$-chain $\delta$ such that $\partial \delta = \alpha + \beta - \alpha\beta$.

Think about $\Delta^2$ as a triangle, and consider a first map $f:\Delta^2\to [0,1]$ that crushes one of the sides against the other two. For a rigourous formulation of this map, consider $\Delta^2 = \{(x_0,x_1,x_2) \mid x_i\geq 0, \sum_i x_i = 1\}$ and a first $g(x_0,x_1,x_2) = \left\{\begin{array}{lr} (x_0-x_2, x_1+2x_2, 0), & \text{if } x_0\geq x_2\\ (0,x_1+2x_0, x_2-x_0), & \text{otherwise } \end{array}\right\}$

an then $l: [e_0, e_1]\cup [e_1,e_2]\to [0,1]$ (where $e_0=(1,0,0)$ etc.) the only " linear deparametrization" (that is, the converse of the only linear parametrization of this union of segments - if you wish I can also write down a formula for this one (linear is not the right word, but I assume you understand what I mean)).

Then $l\circ g: \Delta^2\to [0,1]$ is the $f$ I'm talking about. Then $(\alpha\beta)\circ f$ is our $\delta$ (maybe up to a minus sign, but it doesn't really matter).

Indeed, $\partial \delta = \delta\circ l_{12} - \delta\circ l_{02} + \delta\circ l_{01}$ where $l_{ij}$ is the unique linear parametrization of $[e_i, e_j]$ by $[0,1]$.

Let's now check that this is what we want: for instance $\delta\circ l_{01} = (\alpha\beta)\circ f \circ l_{01} = (\alpha\beta)\circ l\circ g \circ l_{01}$. But notice that $g$ is the identity on $[e_0,e_1]$, so $\delta\circ l_{01}= (\alpha\beta)\circ l\circ l_{01}$. Now given the definitions of $l$ and $l_{01}$ it is clear that $l\circ l_{01}$ is precisely $[0,1]\to [0,\frac{1}{2}]$, $t\mapsto \frac{t}{2}$ (again, we may write down the formulas to prove this explicitly but if you look geometrically at how everything was defined, it should be clear), so that $(\alpha\beta)\circ l\circ l_{01} = \alpha$

Similarly, one may check that $\delta\circ l_{12} = \beta$ and $\delta\circ l_{02}$ is $\alpha\beta$ (though that last one may require more formulas to see clearly - but the geometric picture should leave you no doubt about this).

Hence we have our lemma. Thus $(\gamma\tau)\gamma^{-1}$ is in particular homologous to $\gamma + \tau + \gamma^{-1}$ (I've added parentheses because at first it may not be clear that homotopic paths are homologous)

It remains to check that $\gamma^{-1}$ is homologous to $-\gamma$, and this will complete the proof.

But we know that $\gamma\gamma^{-1}$ is homologous to $\gamma+\gamma^{-1}$ so it suffices to show that $\gamma\gamma^{-1}$ is homologous to zero to conclude. But $\gamma\gamma^{-1}$ is path-homotopic to a constant path, so there are just two things that now suffice to conclude :

A constant path is homologous to zero

Two path-homotopic paths are homologous

Let $c$ be a constant path, say equal to $x$. Then $c$ is the boundary of the constant $2$-simplex $\tilde{c}: \Delta^2\to X$, constant equal to $x$: hence $c$ is homologous to zero.

The second statement is the more interesting one.

So let $\alpha, \beta$ be two homotopic paths, and let $H$ be a path-homotopy between them, $H(-,0) =\alpha, H(-,1) =\beta$

The explicit formulas are annoying to write, once again I may write them if you wish; but for starters let's try to get a geometric picture: $[0,1]\times [0,1]$ is like a square; we may write it as two triangles $T_1, T_2$ whose intersection is precisely the diagonal of the square. If you now consider these triangles as simplices $t_i : \Delta^2 \to T_i$, then one may try to compute $\partial t_1,\partial t_2$. One notices soon enough that (there may be differences according to how one chooses $t_1,t_2$ exactly) $\partial t_1 = $ a constant $1$-simplex $-$ the diagonal $1$-simplex $+\alpha$ and something similar for $\beta$. But the diagonal is the same in both cases (provided we chose $t_1,t_2$ coherently), so we may now substract $\partial t_1 - \partial t_2$ and get $\alpha - \beta +$ some constant paths. We already constant paths are homologous to zero, so $\alpha-\beta$ is as well, so $\alpha$ and $\beta$ are homologous.

This shows how one may relate homotopy of paths and homology; in particular this is the beginning of Hurewicz's theorem.

I hope this answers your question (and maybe does a bit more)

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    $\begingroup$ Thank you so much, extremely clear! $\endgroup$ – B.nia Jul 2 '18 at 18:53

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